Let
x and
y be any two elements in the domain
(Z), such that
f(x+y)=f(x)+f(y)...(i)
Differentiating above expression w.r.t '
y, keeping
x constant, we get
f′(x+y)=f(y) Let
y=0⇒f′(x+0)=f′(0) and Assume
f′(0)=K ∴f′(x)=K Integrating on both sides, we get
f(x)=Kx+C{C→integration constant}....(ii)
Now putting
x=y=0 in Eq. (i), we get
f(0+0)=f(0)+f(0)⇒f(0)=2f(0) or f(0)=0
Let
x=0, from Eq. (ii)
f(0)=K×0+C or
0=0+C or
C=0 ∴f(x)=kx Case (i) If
k>0, then
f(x) is strictly increasing.
Case (ii) If
k<0, then
f(x) is strictly decreasing
So, function is injective
Also
f(x) where every element in the codomain is a valid output of the function i.e., range is equal to codomain.
So, function is surjective also.
Therefore, there are two bijective functions.